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Math Help - Differentiation using first principles

  1. #1
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    Differentiation using first principles

    I'm currently studying a chapter on Differentiation. I was introduced to first prinicples, and i found a few of the first questions easy. But, ive hit a dead end, and nothing I try with this next question works. I looked up the method and found explanations at:
    First Principles
    However, this does not help me get the answer I seek. Could you please help?

    The question goes:

    "Use First principles to find dy/dx

    y=6x-2x^2

    Now i used the formula, and got something like this:

    \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}

    = \lim_{h \rightarrow 0} \frac{6(x+h)-2(x+h)^2-(6x-2x^2)}{h}

    From here, I have no idea, Ive tried many different ways and I always end up with the wrong answer.

    Please explain to me how to do this.
    Last edited by mr fantastic; October 15th 2009 at 03:29 AM. Reason: Improved latex code
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  2. #2
    Junior Member enjam's Avatar
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    Ok, so we have y = -2x^{2} + 6x, so
    f(x+h) = -2(x+h)^{2} + 6(x+h)
    we'll expand this, such that we get:
    -2x{^2} -4xh - 2h{^2} + 6x + 6h

    We now need to find:
    dy/dx = \lim_{h->0} [f(x+h) - f(x)] \div (h)

    = \lim_{h->0} [(-2x^{2} - 4xh - 2x^{2} + 6x + 6h) - (-2x^{2} + 6x)] \div (h)

    = \lim_{h->0} h(-2h - 4x + 6)\div h

    = \lim_{h->0} (-2h -4x + 6)

    = -4x + 6
    Last edited by enjam; October 15th 2009 at 03:19 AM.
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  3. #3
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    Quote Originally Posted by enjam View Post
    = \lim_{h->0} [(-2x^{2} - 4xh - 2x^{2} + 6x + 6h) - (-2x^{2} + 6x)] \div (h)

    = \lim_{h->0} h(-2h - 4x + 6)\div h
    Could you please tell me how you get from the first part of this to the second?
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  4. #4
    Junior Member enjam's Avatar
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    Sorry, about that made a small typo, it's actually meant to be this:
    \lim_{h->0} [(-2x^{2} - 4xh - 2h^{2} + 6x + 6h) - (-2x^{2} + 6x)] \div (h)

    So we have -2x^2 + 2x^2 + 6x - 6x - 4xh - 2h^2 + 6h

    = -4xh -2h^2 + 6h
    = h(-4x - 2h + 6)
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